JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
A plane electromagnetic wave of wavelength \(\lambda \) has an intensity \(I.\) It is propagating along the positive \(Y-\) direction. The allowed expressions for the electric and magnetic fields are given by
- A \(\vec E\, = \,\sqrt {\frac{I}{{{\varepsilon _0}C}}} \cos \left[ {\frac{{2\pi }}{\lambda }(y - ct)} \right]\,\hat i\,;\,\vec B\, = \,\frac{1}{c}E\hat k\)
- B \(\vec E\, = \,\sqrt {\frac{I}{{{\varepsilon _0}C}}} \cos \left[ {\frac{{2\pi }}{\lambda }(y - ct)} \right]\,\hat k\,;\,\vec B\, = - \,\frac{1}{c}E\hat i\)
- C \(\vec E\, = \,\sqrt {\frac{{2I}}{{{\varepsilon _0}C}}} \cos \left[ {\frac{{2\pi }}{\lambda }(y - ct)} \right]\,\hat k\,;\,\vec B\, = + \frac{1}{c}E\hat i\)
- D \(\vec E\, = \,\sqrt {\frac{{2I}}{{{\varepsilon _0}C}}} \cos \left[ {\frac{{2\pi }}{\lambda }(y + ct)} \right]\,\hat k\,;\,\vec B\, = \frac{1}{c}E\hat i\)
Answer & Solution
Correct Answer
(C) \(\vec E\, = \,\sqrt {\frac{{2I}}{{{\varepsilon _0}C}}} \cos \left[ {\frac{{2\pi }}{\lambda }(y - ct)} \right]\,\hat k\,;\,\vec B\, = + \frac{1}{c}E\hat i\)
Step-by-step Solution
Detailed explanation
If \(E_0\) is magnitude of electric field then \(\frac{1}{2}\,{\varepsilon _0}{E_0}^2\, \times \,C\, = \,I\, \Rightarrow {\kern 1pt} {E_0}\, = \sqrt {\frac{{2I}}{{C{\varepsilon _0}}}} \) \({B_0} = {\kern 1pt} \frac{{{E_0}}}{C}\,\) Direction of \(\vec E \times \vec B\) will be…
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